Complete the square to solve for $x$. $x^{2}-12x+27 = 0$
Answer: Begin by moving the constant term to the right side of the equation. $x^2 - 12x = -27$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-12$ , half of it would be $-6$ , and squaring it gives us ${36}$ $x^2 - 12x { + 36} = -27 { + 36}$ We can now rewrite the left side of the equation as a squared term. $( x - 6 )^2 = 9$ Take the square root of both sides. $x - 6 = \pm3$ Isolate $x$ to find the solution(s). $x = 6\pm3$ So the solutions are: $x = 9 \text{ or } x = 3$ We already found the completed square: $( x - 6 )^2 = 9$